Q.Integrate the following function: $\sqrt{x^2+4x+6}$

Idea: complete the square, then apply the standard $\int\sqrt{u^2+a^2}\,du$ formula.

Standard result:

Here $\dfrac{a^2}{2}=\dfrac{2}{2}=1$, so substituting back $u=x+2$:

Step 1 — Complete the square

Half of the middle coefficient $4$ is $2$, and $(x+2)^2=x^2+4x+4$, so

The integral becomes $\displaystyle\int\sqrt{(x+2)^2+2}\,dx$, of the form $\sqrt{u^2+a^2}$ with $u=x+2$ and $a=\sqrt2$ (so $a^2=2$).

Step 2 — The standard formula

With $a^2=2$, the log coefficient is $\dfrac{a^2}{2}=\dfrac{2}{2}=1$ — not $\tfrac12$. So

Step 3 — Substitute back

Replace $u=x+2$ and note $(x+2)^2+2=x^2+4x+6$:

The absolute value matters: the radical is always positive (discriminant $16-24<0$), but $x+2$ can be negative, so the log argument needs $|\cdot|$.

Method: Complete the square, then use a standard $\sqrt{\ }$ formula

To integrate $\sqrt{\text{quadratic}}$, rewrite the quadratic as $(x+p)^{2}\pm a^{2}$ or $a^{2}-(x+p)^{2}$ by completing the square, substitute $t=x+p$, and quote the matching standard integral.

Steps

Step 1: Complete the square on the quadratic under the root, so it becomes $(x+p)^{2}+k$ for some constant $k$.

Step 2: Substitute $t=x+p$ (so $dt=dx$); the integral becomes $\int\sqrt{t^{2}\pm a^{2}}\,dt$ or $\int\sqrt{a^{2}-t^{2}}\,dt$.

Step 3: Apply the correct standard formula.

Step 4: Back-substitute $t=x+p$ and simplify; keep $C$. The whole skill is matching the completed square to the right one of these three templates.

Common Mistakes

Mistake 1: Completing the square wrongly: $x^{2}+4x+6\neq(x+2)^{2}+6$.

Why it's wrong: $(x+2)^{2}=x^{2}+4x+4$, so you must subtract the $4$: $x^{2}+4x+6=(x+2)^{2}+2$. Correct approach: add and subtract $(b/2)^{2}$.

Mistake 2: Using the $\sqrt{a^{2}-t^{2}}$ (arcsin) formula for a $+$ quadratic.

Why it's wrong: $(x+2)^{2}+2$ is a $t^{2}+a^{2}$ form, giving a $\log$, not $\sin^{-1}$. Correct approach: match the sign — a plus constant means the logarithmic template.

Mistake 3: Taking $\dfrac{a^{2}}{2}$ as $\dfrac{a}{2}$ (here $a^{2}=2$).

Why it's wrong: the coefficient is $\dfrac{a^{2}}{2}=1$, not $\dfrac{\sqrt2}{2}$. Correct approach: use $a^{2}$, the constant itself, in the formula.